Calc 3- particle moves in a circle 6 cm............ find parametization and speed. 10 POINTS BEST ANSWER

A particle moves on a circle of radius ...... Calc 3- velocity, motion, speed- 10 POINTS BEST ANSWER?
A particle moves on a circle of radius 6 cm, centered at the origin, in the xy-plane (x and y measured in centimeters). It starts at the point (0,6) and moves counterclockwise, going once around the circle in 10 seconds.

(a) Write a parameterization for the particle's motion.
x(t)=___
y(t)=___

(b) What is the particle's speed? (give units)

speed=___

please include work and answer.

thanks so much


what i did
the speed is 1.2pi cm/s which is write but i still cant get part a

im thinking in has something to do with it going counterclockwise
or because it says it starts at (0,6)

i thought it was
x(t)=-6sint
y(t)=6cost

which would make it start at (0,6) and also make it go counter clock wise.

i think its right but i need to add something because the site is telling me

"Check that your parameterization goes around the circle in the correct amount of time."
so i must have to adjust t or something

ive tried x=6cost
y=sint, ive tried flipping and flipping the signs

i think i need to add the speed or something

The period is 10 seconds. We need a periodic function whose period is 10. Consider sin(t). This is periodic with period 2pi. Moreover, sin(2pi*t) has period 1, and for period 10 it's sin(2pi*t/10)=sin(pi*t/5).

Now, since the radius is 6, we can parametrize the function as:

x(t)=6*cos(pi*t/5)
y(t)=6*sin(pi*t/5)

Now, this would be perfect if the particle had started at (6,0). But it's been rotated 90 degrees. Since cos(x+pi/2)=-sinx and sin(x+pi/2)=cosx, our actual parametrization would be:

x(t)=-6sin(pi*t/5)
y(t)=6cos(pi*t/5)
The speed is 6*(pi/5).

You could use speed. The linear speed divided by the radius gives you the angular speed in radians per unit time. That works, but involves an extra step. You compute the speed v as 2πr/T, where r is the radius and T is period.

It's simpler to note that the particle covers 2π radians in T=10 seconds to get ω = 2π/T as the angular speed in radians/sec. The speed is positive, by convention, because the motion is counter-clockwise. Over a time interval t, the particle moves through and angle θ = ωt = 2πt/T.

That makes the angular position at time t is is θ = φ + ωt, where φ is the angular position at time t=0. Since the particle starts at (6,0), φ = 0 in this problem...but don't forget about it. Finally, the (x,y) position at time t is:

x = r cos θ = r cos (φ + ωt) = 6 cos(2πt/10)
y = r sin θ = r sin (φ + ωt) = 6 sin (2πt/10)

Oh, I didn't look ahead. You do need the linear speed after all, but you already had that. v = 2πr/T (or v = ωr) gives v = 2π*6/10 = 6π/5 cm/s like you said.